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On Monday, a person mailed 8 packages weighing an average [#permalink]

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13 Aug 2012, 06:57

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On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of \(12\frac{3}{8}\) pounds, and on Tuesday, 4 packages weighing an average of \(15\frac{1}{4}\) pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

(A) \(13\frac{1}{3}\)

(B) \(13\frac{13}{16}\)

(C) \(15\frac{1}{2}\)

(D) \(15\frac{15}{16}\)

(E) \(16\frac{1}{2}\)

Practice Questions Question: 16 Page: 154 Difficulty: 600

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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13 Aug 2012, 06:57

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SOLUTION

On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of \(12\frac{3}{8}\) pounds, and on Tuesday, 4 packages weighing an average of \(15\frac{1}{4}\) pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

(A) \(13\frac{1}{3}\)

(B) \(13\frac{13}{16}\)

(C) \(15\frac{1}{2}\)

(D) \(15\frac{15}{16}\)

(E) \(16\frac{1}{2}\)

The total weight of 8 packages is \(8*12\frac{3}{8}=99\) pounds;

The total weight of 4 packages is \(4*15\frac{1}{4}=61\) pounds;

The average weight of all 12 packages is \(\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}\).

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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16 Aug 2012, 11:10

Bunuel wrote:

RESERVED FOR A SOLUTION.

Bunuel, had an off-topic request for you: Could you please post questions from non-OG sources as well? I'm not sure if that might breach a copyright arrangement bsaed on the source you use, and of course your comments on other's questions are supremely valuable for those of us subscribed to your daily updates - but if you could include occasional 700+ non-OG questions, would be much appreciated by your "followers"

_________________

How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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17 Aug 2012, 00:54

Expert's post

SOLUTION

On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of \(12\frac{3}{8}\) pounds, and on Tuesday, 4 packages weighing an average of \(15\frac{1}{4}\) pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

(A) \(13\frac{1}{3}\)

(B) \(13\frac{13}{16}\)

(C) \(15\frac{1}{2}\)

(D) \(15\frac{15}{16}\)

(E) \(16\frac{1}{2}\)

The total weight of 8 packages is \(8*12\frac{3}{8}=99\) pounds;

The total weight of 4 packages is \(4*15\frac{1}{4}=61\) pounds;

The average weight of all 12 packages is \(\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}\).

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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07 Sep 2012, 10:09

There must be something easy that i just don't get for me : 8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?

I guess it must be something different of spelling or something?

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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07 Sep 2012, 10:15

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Gmbrox wrote:

There must be something easy that i just don't get for me : 8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?

I guess it must be something different of spelling or something?

Thanks a lot for your help !

It's not 12 multiplied by 3/8. it's \(12\frac{3}{8}=\frac{12*8+3}{8}=\frac{99}{8}\) (the same way as \(1\frac{1}{2}=\frac{3}{2}\)). _________________

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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07 Sep 2012, 10:20

Thank you a lot bunuel, Ok after reviewing the official book, i now got it, it is a mix number, it does not exists in france so that's why. If anyone has difficulties to understand like me :

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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07 Sep 2012, 11:47

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Since the final average cannot be greater than \(15\frac{1}{4}\), answers C, D and E are out.

We can use the property of weighted averages. \(15\frac{1}{4}=15\frac{2}{8}\), the distance between the two initial averages is almost 3. Since the number of packages are in a ratio of 8:4 = 2:1, the differences between the final average and the initial averages are in a ratio 1:2. So, the distance between \(12\frac{3}{8}\) and the final average is almost 1, close to \(12\frac{3}{8}+1\approx{13}\frac{1}{4}\). The final answer should be close to \(13\frac{1}{4}\).

Answer A. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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26 Oct 2013, 04:38

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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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15 Nov 2014, 02:55

I used ratio of packages, which is 2:1. converted both to the same fractions, so 15 1/4 = 15 2/8

2x(12 3/8) + 1x( 15 2/8) = 24+15+ 6/8+2/8 = 39 and 8/8, 8/8 is also obviously 1. Could also be together 40 but that's not easily divisible with three and you know you're left with a remainder. Instead just: 39/3 + 1/3 = 13 and 1/3

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]

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21 Feb 2016, 12:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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